Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 17 - Section 17.1 - Second-Order Linear Equations - 17.1 Exercise - Page 1160: 26

Answer

The solution of the the given boundary-value problem is $$ y(x)=(\frac{-1}{e^6-1})+(\frac{e^6}{e^6-1})e^{-6x}. $$

Work Step by Step

$$ y^{\prime \prime}+6 y^{\prime }=0, \quad y(0)=1, \quad y(1)=0 $$ The given equation is a homogeneous linear equation. So the characteristic equation of the differential equation is $$ r^{2}+6r =r(r+6)=0 $$ whose roots are $$ r=0, -6 $$ Therefore, the general solution of the given differential equation is $$ y(x)=c_{1} +c_{2}e^{-6x}, $$ Then $$ 1=y(0)=c_{1} +c_{2}e^{0} \Rightarrow c_{1} +c_{2}=1 $$ and $$ 0=y(1)=c_{1} +c_{2}e^{-6} \Rightarrow c_{1} +c_{2}e^{-6}=0 $$ so $$ c_{2}=\frac{e^6}{e^6-1},\:c_{1}=1-\frac{e^6}{e^6-1}=\frac{-1}{e^6-1} $$ Thus, the solution of the boundary-value problem is $$ y(x)=(\frac{-1}{e^6-1})+(\frac{e^6}{e^6-1})e^{-6x}. $$
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